A 0.20 M sulfuric acid (H2SO4) solution is titrated with 0.25 M sodium hydroxide (NaOH). Calculate the volume of NaOH required to neutralize 25.0 mL of the H2SO4 solution.
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To calculate the volume of NaOH required to neutralize 25.0 mL of the H2SO4 solution, we need to determine the number of moles of H2SO4 present in the 25.0 mL solution and then use the stoichiometry of the reaction to determine the moles of NaOH required.
First, calculate the moles of H2SO4:
moles H2SO4 = Molarity × Volume
moles H2SO4 = 0.20 M × (25.0 mL/1000 mL)
moles H2SO4 = 0.005 mol
The balanced equation for the reaction between H2SO4 and NaOH is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
According to the stoichiometry of the reaction, 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, the moles of NaOH required to neutralize the H2SO4 is twice the moles of H2SO4:
moles NaOH = 2 × moles H2SO4
moles NaOH = 2 × 0.005 mol
moles NaOH = 0.01 mol
Finally, calculate the volume of NaOH required using the moles and molarity of the NaOH solution:
Volume NaOH = moles NaOH / Molarity NaOH
Volume NaOH = 0.01 mol / 0.25 M
Volume NaOH = 0.04 L
Volume NaOH = 40 mL
Thus, 40 mL of the 0.25 M NaOH solution is required to neutralize 25.0 mL of the 0.20 M H2SO4 solution.
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